Question about area on a sphere

[Vorde]

For a end project in a math class, I am going to attempt to figure out the laws for spherical trigonometry. Specificially, side lengths and areas for triangles on a sphere.

I am fairly sure I know how to go about side lengths, and I have an idea how to go about the area of a curved triangle, but it's dependent on an assumption I want to make sure is correct.

Without going into details, the assumption is that increasing the interior angle sum (past 180 degrees) will increase the area by an amount dependent only on the angle increase (and the dimensions of the triangle), meaning it doesn't matter which angle is being increased, just that the total angle is being increased.

Is this correct? I would appreciate non specifics if it is wrong- I want to try to figure it out by my self.

Thank you.

 

[vkash]

For a end project in a math class, I am going to attempt to figure out the laws for spherical trigonometry. Specificially, side lengths and areas for triangles on a sphere.

I am fairly sure I know how to go about side lengths, and I have an idea how to go about the area of a curved triangle, but it's dependent on an assumption I want to make sure is correct.

Without going into details, the assumption is that increasing the interior angle sum (past 180 degrees) will increase the area by an amount dependent only on the angle increase (and the dimensions of the triangle), meaning it doesn't matter which angle is being increased, just that the total angle is being increased.

Is this correct? I would appreciate non specifics if it is wrong- I want to try to figure it out by my self.

Thank you.

i didn't understand what you say..
How you can increase sum of angles of triangle?

 

[DaveC426913]

i didn't understand what you say..
How you can increase sum of angles of triangle?

...laws for spherical trigonometry.

By putting it on a spherical geometry. It will be greater than 180. More curvature, and that number can go all the way up to 360.

 

[vkash]

By putting it on a spherical geometry. It will be greater than 180. More curvature, and that number can go all the way up to 360.

Oh sorry Vorde it is out of my scope...

 

[DaveC426913]

Just doing some sketching. What is the largest possible triangle on a sphere? What does it look like?

 

[Vorde]

I think it depends on whether or not you allow overlap (going around more than once), if so infinite, if not then my quick guess would be the maximum angle measure is 540 degrees.

I don't think that effects my question though. Also I apologize, it should be sphere not circle in the title but it's too late to change it.

 

[micromass]

I don't think that effects my question though. Also I apologize, it should be sphere not circle in the title but it's too late to change it.

I changed it for you.

Anyway, here is a good free book on spherical trig: http://www.gutenberg.org/ebooks/19770

 

[mathwonk]

First figure out the relation between angles and areas of "lunes", which is the region between two great circles. Then try to bootstrap up to a triangle, which after all is an intersection of lunes. for simplicity start with a sphere of radius one. you might also stick to triangles lying in one hemisphere. i.e. just as a triangle in the plane is a bounded intersection of three half planes, on the sphere one can define it as the intersection of three hemispheres.

 

[DaveC426913]

I think it depends on whether or not you allow overlap (going around more than once), if so infinite, if not then my quick guess would be the maximum angle measure is 540 degrees.

I would disallow overlap. Likewise, if we allowed overlap when defining regular polyhedra, there could be a lot more than the basic 5 platonic solids.

I figured the largest angle would be 360 degrees. All three vertices would be coincident on the "back" of the sphere, and the sum angle would be 360. i.e. a point - er... - everything except a point.

When I sketch it out, it makes no sense. The largest triangle on a sphere ends up having its 3 vertices coincident, sides of length zero, and area equal to the area of the sphere.

i.e. asking what the largest triangle on a spherical surface could be is equivalent to asking for the inverse of what the smallest area on a plane surface is (which is zero).

See my back-of-napkin sketch.

 

[Vorde]

I changed it for you.

Anyway, here is a good free book on spherical trig: http://www.gutenberg.org/ebooks/19770

Thank you on both counts, I am hesitant to follow the link simply because I don't want to accidentally read something I'm trying to figure out on my own, but in the end I might have to.

First figure out the relation between angles and areas of "lunes", which is the region between two great circles. Then try to bootstrap up to a triangle, which after all is an intersection of lunes. for simplicity start with a sphere of radius one. you might also stick to triangles lying in one hemisphere. i.e. just as a triangle in the plane is a bounded intersection of three half planes, on the sphere one can define it as the intersection of three hemispheres.

Good advice, thanks. I actually think I have the beginnings of an idea following this path.

I would disallow overlap. Likewise, if we allowed overlap when defining regular polyhedra, there could be a lot more than the basic 5 platonic solids.

I figured the largest angle would be 360 degrees. All three vertices would be coincident on the "back" of the sphere, and the sum angle would be 360. i.e. a point - er... - everything except a point.

When I sketch it out, it makes no sense. The largest triangle on a sphere ends up having its 3 vertices coincident, sides of length zero, and area equal to the area of the sphere.

i.e. asking what the largest triangle on a spherical surface could be is equivalent to asking for the inverse of what the smallest area on a plane surface is (which is zero).

See my back-of-napkin sketch.

Here is what I was thinking when I said 540 degrees. You could have an isosceles triangle with one point at the pole and two on the equator, the angle value for each point on the equator would be 90, and the angle value at the pole would be dependent on the distance between the two equator points. By varying that distance, and not letting the points overlap, you could get the angle of the pole point up to 360 (or infinitely close to that).

 

[DaveC426913]

Here is what I was thinking when I said 540 degrees. You could have an isosceles triangle with one point at the pole and two on the equator, the angle value for each point on the equator would be 90, and the angle value at the pole would be dependent on the distance between the two equator points. By varying that distance, and not letting the points overlap, you could get the angle of the pole point up to 360 (or infinitely close to that).

Huh. You're right.

In fact, it goes even further. You could then move the equatorial line segment south toward the south pole, which would increase the two equatorial angles equally. In fact, you could bring it all the way around until you had the entire sphere inside the triangle except for an infinitesimally small triangle outside it. And that means that the sum of the angles of the triangle are 3 x (360-60) = 900 degrees total!

The only question remaining being this: are we really measuring the interior angles of the triangle anymore?

It is tantamount to counting the exterior angles of a triangle in Cartesian space and claiming the triangle contains the entire universe except what is inside this little area.

 

[Vorde]

If you define the inside of the triangle to be any point where a line extending in any direction from that point will intersect one of the sides of the triangle, then yes I would say it is still the interior, although it doesn't seem it anymore.

However I haven't convinced myself that moving the equatorial points towards the south pole will increase the interior angle past 540. I'm having a tough time visualizing this though. It makes sense that a triangle like I described in my last post would account for half the surface area of the sphere, all you have to do to make the triangle as large as possible is to lower the equatorial points in that hemisphere triangle down to the south pole. It seems to me that while doing this the interior angle of the equatorial points wouldn't change, but as I said I can't visualize this well so I may well be mistaken.

 

[mathwonk]

if you accept my suggestion restrict to intersections of three hemispheres, you can still get a triangle with all three vertices arbitrarily close to the equator (and the triangle would fill up more and more of a hemisphere). Then what would the three angles approach?

 

[DaveC426913]

if you accept my suggestion restrict to intersections of three hemispheres,

Not sure what this phrase means. Do you mean only use world lines?

 

[AlephZero]

The only question remaining being this: are we really measuring the interior angles of the triangle anymore?

It is tantamount to counting the exterior angles of a triangle in Cartesian space and claiming the triangle contains the entire universe except what is inside this little area.

But the surface of a sphere isn't a plane and isn't infinite, so the "common sense" analogies doesn't take the math forward much.

If you look at Euclid's Elements, the Greeks apparently never noticed there was an issue here. Euclid takes the idea of two points being on the same side or on opposite sides of a line as "obvious". There is a fairly well-known "proof" that all triangles are isosceles, based on drawing an "impossible" figure where two lines interect on the wrong side of a third line.

The OP will have to make some definitions about what is "valid", and figure out the consequences. It might be worth a look at the start of Hilbert's book "Foundations of Geometry" where he takes Euclid's axioms apart and puts them back together again, for plane geometry. It's a very readable book - as well as being a great mathematician, Hilbert was pretty good at explaining things. It's on Project Gutenberg, and elswhere on the web in Gutenberg doesn't have the diagrams.

 

[Vorde]

Looking back, although my question wasn't directly answered, I think (thanks to you all) that I have a better way of solving my problem now.

Thank you all for your help.

 

[mathwonk]

the intersection of hemispheres is always contained in a hemisphere, so is more restrictive than a triangle whose sides are geodesics.

 

[Mholnic-]

If you define the inside of the triangle to be any point where a line extending in any direction from that point will intersect one of the sides of the triangle, then yes I would say it is still the interior, although it doesn't seem it anymore.

At risk of butting into a conversation not my own just to satisfy my own curiosity.. wouldn't the above statement be true for ALL triangles existing on a sphere? Including the "infinitesimally small triangle" that DaveC~ mentioned on the 'outside' of your 900~ degree triangle. On an infinite plane you could test whether you were inside or outside the triangle, but on a finite sphere... I don't believe you can. By creating one triangle on a spherical surface you automatically create a second triangle whose angles are complimentary to the other (outside or inside) triangle's angles; given this, wouldn't "inside" and "outside" just be arbitrary definitions based on personal preference?

 

[Vorde]

I have to think about this more (referring to Mholnic's conversation- only tangent to the original question), I'm having doubts about some of the triangles discussed before, specifically relating to how great circles only line up to lat/long lines once on a sphere.

Would someone with better knowledge confirm this for me: you can't take a equatorial circle and reduce it in size to a point (at the pole) whilst keeping it a geodesic.

I believe that statement is correct, and if so I believe I was originally correct by saying that the maximum angle measure for a spherical triangle is 540.

 

[DaveC426913]

At risk of butting into a conversation not my own just to satisfy my own curiosity.. wouldn't the above statement be true for ALL triangles existing on a sphere? Including the "infinitesimally small triangle" that DaveC~ mentioned on the 'outside' of your 900~ degree triangle. On an infinite plane you could test whether you were inside or outside the triangle, but on a finite sphere... I don't believe you can. By creating one triangle on a spherical surface you automatically create a second triangle whose angles are complimentary to the other (outside or inside) triangle's angles; given this, wouldn't "inside" and "outside" just be arbitrary definitions based on personal preference?

Wot he sed.

 

[mathwonk]

If you define a triangle, together with its interior, as the intersection of three distinct hemispheres whose boundary great circles have no common point, then a triangle is necessarily contained in a hemisphere, and the angle sum of the interior angles, which is well defined here, is always less than 540 degrees. This is also the least upper bound, but is not attained.

If you define a triangle merely as determined by three non collinear points, together with three geodesics joining those points in pairs, plus a choice of inside, then indeed each triple of such geodesics determines two triangles, and by taking one of them very small, with angle sum near 180, the other will have angle sum as near 900 as desired.

If the radius of the sphere is one, then the angle sum, in radians, equals the area plus pi. Hence for a more general sphere, the angle sum minus pi, the "angle excess" of a triangle, equals the area of the triangle times the curvature of the sphere, where the curvature is the reciprocal of the square of the radius.

Note this holds also for both geodesic triangles determined by a given triple of geodesic arcs, since the sum of the angle excesses for a triangle and its complementary triangle, equals 4pi, i.e. the area of the sphere times its curvature.

Indeed the same formula holds also for Euclidean and hyperbolic triangles, where in those cases the curvature is either zero or negative.

Indeed on any curved surface, the integral of the curvature over a geodesic triangle equals the angle excess. if we triangulate the surface by geodesic triangles, with V vertices, E edges, and F faces, and note that the total angle sum is 2piV, then the total angle excess is 2piV-piF. Since also 2E = 3F, it follows that the integral of the curvature over the surface is 2pi(V-E+F), which is called the Gauss Bonnet theorem.

 

[DaveC426913]

If you define a triangle, together with its interior, as the intersection of three distinct hemispheres whose boundary great circles have no common point,

Colour me dense, I just don't get this. A hemisphere is bounded by a great circle. How can you have two great circles that do not intersect (have no common point), let alone three? Any two great circles will intersect at two points. Am I misunderstanding of the term 'common point'?

 

[mathwonk]

there are three boundary circles but no point lies on all three. Thus the three great circles form 8 triangles, but by choosing one hemisphere bounded by each circle, and intersecting these hemispheres, we have chosen one of the 8 triangles, along with its interior.